3.35 \(\int \frac {x \cosh (c+d x)}{(a+b x)^3} \, dx\)

Optimal. Leaf size=178 \[ -\frac {a d^2 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{2 b^4}-\frac {a d^2 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{2 b^4}+\frac {d \sinh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{b^3}+\frac {d \cosh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{b^3}+\frac {a d \sinh (c+d x)}{2 b^3 (a+b x)}-\frac {\cosh (c+d x)}{b^2 (a+b x)}+\frac {a \cosh (c+d x)}{2 b^2 (a+b x)^2} \]

[Out]

-1/2*a*d^2*Chi(a*d/b+d*x)*cosh(-c+a*d/b)/b^4+1/2*a*cosh(d*x+c)/b^2/(b*x+a)^2-cosh(d*x+c)/b^2/(b*x+a)+d*cosh(-c
+a*d/b)*Shi(a*d/b+d*x)/b^3-d*Chi(a*d/b+d*x)*sinh(-c+a*d/b)/b^3+1/2*a*d^2*Shi(a*d/b+d*x)*sinh(-c+a*d/b)/b^4+1/2
*a*d*sinh(d*x+c)/b^3/(b*x+a)

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Rubi [A]  time = 0.37, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6742, 3297, 3303, 3298, 3301} \[ -\frac {a d^2 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{2 b^4}+\frac {d \sinh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{b^3}-\frac {a d^2 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{2 b^4}+\frac {d \cosh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{b^3}+\frac {a d \sinh (c+d x)}{2 b^3 (a+b x)}-\frac {\cosh (c+d x)}{b^2 (a+b x)}+\frac {a \cosh (c+d x)}{2 b^2 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*Cosh[c + d*x])/(a + b*x)^3,x]

[Out]

(a*Cosh[c + d*x])/(2*b^2*(a + b*x)^2) - Cosh[c + d*x]/(b^2*(a + b*x)) - (a*d^2*Cosh[c - (a*d)/b]*CoshIntegral[
(a*d)/b + d*x])/(2*b^4) + (d*CoshIntegral[(a*d)/b + d*x]*Sinh[c - (a*d)/b])/b^3 + (a*d*Sinh[c + d*x])/(2*b^3*(
a + b*x)) + (d*Cosh[c - (a*d)/b]*SinhIntegral[(a*d)/b + d*x])/b^3 - (a*d^2*Sinh[c - (a*d)/b]*SinhIntegral[(a*d
)/b + d*x])/(2*b^4)

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {x \cosh (c+d x)}{(a+b x)^3} \, dx &=\int \left (-\frac {a \cosh (c+d x)}{b (a+b x)^3}+\frac {\cosh (c+d x)}{b (a+b x)^2}\right ) \, dx\\ &=\frac {\int \frac {\cosh (c+d x)}{(a+b x)^2} \, dx}{b}-\frac {a \int \frac {\cosh (c+d x)}{(a+b x)^3} \, dx}{b}\\ &=\frac {a \cosh (c+d x)}{2 b^2 (a+b x)^2}-\frac {\cosh (c+d x)}{b^2 (a+b x)}+\frac {d \int \frac {\sinh (c+d x)}{a+b x} \, dx}{b^2}-\frac {(a d) \int \frac {\sinh (c+d x)}{(a+b x)^2} \, dx}{2 b^2}\\ &=\frac {a \cosh (c+d x)}{2 b^2 (a+b x)^2}-\frac {\cosh (c+d x)}{b^2 (a+b x)}+\frac {a d \sinh (c+d x)}{2 b^3 (a+b x)}-\frac {\left (a d^2\right ) \int \frac {\cosh (c+d x)}{a+b x} \, dx}{2 b^3}+\frac {\left (d \cosh \left (c-\frac {a d}{b}\right )\right ) \int \frac {\sinh \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^2}+\frac {\left (d \sinh \left (c-\frac {a d}{b}\right )\right ) \int \frac {\cosh \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^2}\\ &=\frac {a \cosh (c+d x)}{2 b^2 (a+b x)^2}-\frac {\cosh (c+d x)}{b^2 (a+b x)}+\frac {d \text {Chi}\left (\frac {a d}{b}+d x\right ) \sinh \left (c-\frac {a d}{b}\right )}{b^3}+\frac {a d \sinh (c+d x)}{2 b^3 (a+b x)}+\frac {d \cosh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (\frac {a d}{b}+d x\right )}{b^3}-\frac {\left (a d^2 \cosh \left (c-\frac {a d}{b}\right )\right ) \int \frac {\cosh \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{2 b^3}-\frac {\left (a d^2 \sinh \left (c-\frac {a d}{b}\right )\right ) \int \frac {\sinh \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{2 b^3}\\ &=\frac {a \cosh (c+d x)}{2 b^2 (a+b x)^2}-\frac {\cosh (c+d x)}{b^2 (a+b x)}-\frac {a d^2 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (\frac {a d}{b}+d x\right )}{2 b^4}+\frac {d \text {Chi}\left (\frac {a d}{b}+d x\right ) \sinh \left (c-\frac {a d}{b}\right )}{b^3}+\frac {a d \sinh (c+d x)}{2 b^3 (a+b x)}+\frac {d \cosh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (\frac {a d}{b}+d x\right )}{b^3}-\frac {a d^2 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (\frac {a d}{b}+d x\right )}{2 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 158, normalized size = 0.89 \[ -\frac {d (a+b x)^2 \left (\text {Chi}\left (d \left (\frac {a}{b}+x\right )\right ) \left (a d \cosh \left (c-\frac {a d}{b}\right )-2 b \sinh \left (c-\frac {a d}{b}\right )\right )+\text {Shi}\left (d \left (\frac {a}{b}+x\right )\right ) \left (a d \sinh \left (c-\frac {a d}{b}\right )-2 b \cosh \left (c-\frac {a d}{b}\right )\right )\right )+b \cosh (d x) (b \cosh (c) (a+2 b x)-a d \sinh (c) (a+b x))-b \sinh (d x) (a d \cosh (c) (a+b x)-b \sinh (c) (a+2 b x))}{2 b^4 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Cosh[c + d*x])/(a + b*x)^3,x]

[Out]

-1/2*(b*Cosh[d*x]*(b*(a + 2*b*x)*Cosh[c] - a*d*(a + b*x)*Sinh[c]) - b*(a*d*(a + b*x)*Cosh[c] - b*(a + 2*b*x)*S
inh[c])*Sinh[d*x] + d*(a + b*x)^2*(CoshIntegral[d*(a/b + x)]*(a*d*Cosh[c - (a*d)/b] - 2*b*Sinh[c - (a*d)/b]) +
(-2*b*Cosh[c - (a*d)/b] + a*d*Sinh[c - (a*d)/b])*SinhIntegral[d*(a/b + x)]))/(b^4*(a + b*x)^2)

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fricas [B]  time = 0.53, size = 373, normalized size = 2.10 \[ -\frac {2 \, {\left (2 \, b^{3} x + a b^{2}\right )} \cosh \left (d x + c\right ) + {\left ({\left (a^{3} d^{2} - 2 \, a^{2} b d + {\left (a b^{2} d^{2} - 2 \, b^{3} d\right )} x^{2} + 2 \, {\left (a^{2} b d^{2} - 2 \, a b^{2} d\right )} x\right )} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) + {\left (a^{3} d^{2} + 2 \, a^{2} b d + {\left (a b^{2} d^{2} + 2 \, b^{3} d\right )} x^{2} + 2 \, {\left (a^{2} b d^{2} + 2 \, a b^{2} d\right )} x\right )} {\rm Ei}\left (-\frac {b d x + a d}{b}\right )\right )} \cosh \left (-\frac {b c - a d}{b}\right ) - 2 \, {\left (a b^{2} d x + a^{2} b d\right )} \sinh \left (d x + c\right ) - {\left ({\left (a^{3} d^{2} - 2 \, a^{2} b d + {\left (a b^{2} d^{2} - 2 \, b^{3} d\right )} x^{2} + 2 \, {\left (a^{2} b d^{2} - 2 \, a b^{2} d\right )} x\right )} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) - {\left (a^{3} d^{2} + 2 \, a^{2} b d + {\left (a b^{2} d^{2} + 2 \, b^{3} d\right )} x^{2} + 2 \, {\left (a^{2} b d^{2} + 2 \, a b^{2} d\right )} x\right )} {\rm Ei}\left (-\frac {b d x + a d}{b}\right )\right )} \sinh \left (-\frac {b c - a d}{b}\right )}{4 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(d*x+c)/(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/4*(2*(2*b^3*x + a*b^2)*cosh(d*x + c) + ((a^3*d^2 - 2*a^2*b*d + (a*b^2*d^2 - 2*b^3*d)*x^2 + 2*(a^2*b*d^2 - 2
*a*b^2*d)*x)*Ei((b*d*x + a*d)/b) + (a^3*d^2 + 2*a^2*b*d + (a*b^2*d^2 + 2*b^3*d)*x^2 + 2*(a^2*b*d^2 + 2*a*b^2*d
)*x)*Ei(-(b*d*x + a*d)/b))*cosh(-(b*c - a*d)/b) - 2*(a*b^2*d*x + a^2*b*d)*sinh(d*x + c) - ((a^3*d^2 - 2*a^2*b*
d + (a*b^2*d^2 - 2*b^3*d)*x^2 + 2*(a^2*b*d^2 - 2*a*b^2*d)*x)*Ei((b*d*x + a*d)/b) - (a^3*d^2 + 2*a^2*b*d + (a*b
^2*d^2 + 2*b^3*d)*x^2 + 2*(a^2*b*d^2 + 2*a*b^2*d)*x)*Ei(-(b*d*x + a*d)/b))*sinh(-(b*c - a*d)/b))/(b^6*x^2 + 2*
a*b^5*x + a^2*b^4)

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giac [B]  time = 0.14, size = 529, normalized size = 2.97 \[ -\frac {a b^{2} d^{2} x^{2} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} + a b^{2} d^{2} x^{2} {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} + 2 \, a^{2} b d^{2} x {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} - 2 \, b^{3} d x^{2} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} + 2 \, a^{2} b d^{2} x {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} + 2 \, b^{3} d x^{2} {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} + a^{3} d^{2} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} - 4 \, a b^{2} d x {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} + a^{3} d^{2} {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} + 4 \, a b^{2} d x {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} - a b^{2} d x e^{\left (d x + c\right )} + a b^{2} d x e^{\left (-d x - c\right )} - 2 \, a^{2} b d {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} + 2 \, a^{2} b d {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} - a^{2} b d e^{\left (d x + c\right )} + 2 \, b^{3} x e^{\left (d x + c\right )} + a^{2} b d e^{\left (-d x - c\right )} + 2 \, b^{3} x e^{\left (-d x - c\right )} + a b^{2} e^{\left (d x + c\right )} + a b^{2} e^{\left (-d x - c\right )}}{4 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(d*x+c)/(b*x+a)^3,x, algorithm="giac")

[Out]

-1/4*(a*b^2*d^2*x^2*Ei((b*d*x + a*d)/b)*e^(c - a*d/b) + a*b^2*d^2*x^2*Ei(-(b*d*x + a*d)/b)*e^(-c + a*d/b) + 2*
a^2*b*d^2*x*Ei((b*d*x + a*d)/b)*e^(c - a*d/b) - 2*b^3*d*x^2*Ei((b*d*x + a*d)/b)*e^(c - a*d/b) + 2*a^2*b*d^2*x*
Ei(-(b*d*x + a*d)/b)*e^(-c + a*d/b) + 2*b^3*d*x^2*Ei(-(b*d*x + a*d)/b)*e^(-c + a*d/b) + a^3*d^2*Ei((b*d*x + a*
d)/b)*e^(c - a*d/b) - 4*a*b^2*d*x*Ei((b*d*x + a*d)/b)*e^(c - a*d/b) + a^3*d^2*Ei(-(b*d*x + a*d)/b)*e^(-c + a*d
/b) + 4*a*b^2*d*x*Ei(-(b*d*x + a*d)/b)*e^(-c + a*d/b) - a*b^2*d*x*e^(d*x + c) + a*b^2*d*x*e^(-d*x - c) - 2*a^2
*b*d*Ei((b*d*x + a*d)/b)*e^(c - a*d/b) + 2*a^2*b*d*Ei(-(b*d*x + a*d)/b)*e^(-c + a*d/b) - a^2*b*d*e^(d*x + c) +
2*b^3*x*e^(d*x + c) + a^2*b*d*e^(-d*x - c) + 2*b^3*x*e^(-d*x - c) + a*b^2*e^(d*x + c) + a*b^2*e^(-d*x - c))/(
b^6*x^2 + 2*a*b^5*x + a^2*b^4)

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maple [B]  time = 0.11, size = 435, normalized size = 2.44 \[ -\frac {d^{3} {\mathrm e}^{-d x -c} a x}{4 b^{2} \left (b^{2} d^{2} x^{2}+2 a b \,d^{2} x +a^{2} d^{2}\right )}-\frac {d^{3} {\mathrm e}^{-d x -c} a^{2}}{4 b^{3} \left (b^{2} d^{2} x^{2}+2 a b \,d^{2} x +a^{2} d^{2}\right )}-\frac {d^{2} {\mathrm e}^{-d x -c} x}{2 b \left (b^{2} d^{2} x^{2}+2 a b \,d^{2} x +a^{2} d^{2}\right )}-\frac {d^{2} {\mathrm e}^{-d x -c} a}{4 b^{2} \left (b^{2} d^{2} x^{2}+2 a b \,d^{2} x +a^{2} d^{2}\right )}+\frac {d^{2} {\mathrm e}^{\frac {d a -c b}{b}} \Ei \left (1, d x +c +\frac {d a -c b}{b}\right ) a}{4 b^{4}}+\frac {d \,{\mathrm e}^{\frac {d a -c b}{b}} \Ei \left (1, d x +c +\frac {d a -c b}{b}\right )}{2 b^{3}}+\frac {d^{2} {\mathrm e}^{d x +c} a}{4 b^{4} \left (\frac {a d}{b}+d x \right )^{2}}+\frac {d^{2} {\mathrm e}^{d x +c} a}{4 b^{4} \left (\frac {a d}{b}+d x \right )}+\frac {d^{2} {\mathrm e}^{-\frac {d a -c b}{b}} \Ei \left (1, -d x -c -\frac {d a -c b}{b}\right ) a}{4 b^{4}}-\frac {d \,{\mathrm e}^{d x +c}}{2 b^{3} \left (\frac {a d}{b}+d x \right )}-\frac {d \,{\mathrm e}^{-\frac {d a -c b}{b}} \Ei \left (1, -d x -c -\frac {d a -c b}{b}\right )}{2 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(d*x+c)/(b*x+a)^3,x)

[Out]

-1/4*d^3*exp(-d*x-c)/b^2/(b^2*d^2*x^2+2*a*b*d^2*x+a^2*d^2)*a*x-1/4*d^3*exp(-d*x-c)/b^3/(b^2*d^2*x^2+2*a*b*d^2*
x+a^2*d^2)*a^2-1/2*d^2*exp(-d*x-c)/b/(b^2*d^2*x^2+2*a*b*d^2*x+a^2*d^2)*x-1/4*d^2*exp(-d*x-c)/b^2/(b^2*d^2*x^2+
2*a*b*d^2*x+a^2*d^2)*a+1/4*d^2/b^4*exp((a*d-b*c)/b)*Ei(1,d*x+c+(a*d-b*c)/b)*a+1/2*d/b^3*exp((a*d-b*c)/b)*Ei(1,
d*x+c+(a*d-b*c)/b)+1/4*d^2/b^4*exp(d*x+c)/(a*d/b+d*x)^2*a+1/4*d^2/b^4*exp(d*x+c)/(a*d/b+d*x)*a+1/4*d^2/b^4*exp
(-(a*d-b*c)/b)*Ei(1,-d*x-c-(a*d-b*c)/b)*a-1/2*d/b^3*exp(d*x+c)/(a*d/b+d*x)-1/2*d/b^3*exp(-(a*d-b*c)/b)*Ei(1,-d
*x-c-(a*d-b*c)/b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {x e^{\left (d x + c\right )}}{b^{4} d x^{4} + 4 \, a b^{3} d x^{3} + 6 \, a^{2} b^{2} d x^{2} + 4 \, a^{3} b d x + a^{4} d}\,{d x} - b \int \frac {x}{b^{4} d x^{4} e^{\left (d x + c\right )} + 4 \, a b^{3} d x^{3} e^{\left (d x + c\right )} + 6 \, a^{2} b^{2} d x^{2} e^{\left (d x + c\right )} + 4 \, a^{3} b d x e^{\left (d x + c\right )} + a^{4} d e^{\left (d x + c\right )}}\,{d x} + \frac {x e^{\left (d x + 2 \, c\right )} - x e^{\left (-d x\right )}}{2 \, {\left (b^{3} d x^{3} e^{c} + 3 \, a b^{2} d x^{2} e^{c} + 3 \, a^{2} b d x e^{c} + a^{3} d e^{c}\right )}} - \frac {a e^{\left (-c + \frac {a d}{b}\right )} E_{4}\left (\frac {{\left (b x + a\right )} d}{b}\right )}{2 \, {\left (b x + a\right )}^{3} b d} + \frac {a e^{\left (c - \frac {a d}{b}\right )} E_{4}\left (-\frac {{\left (b x + a\right )} d}{b}\right )}{2 \, {\left (b x + a\right )}^{3} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(d*x+c)/(b*x+a)^3,x, algorithm="maxima")

[Out]

b*integrate(x*e^(d*x + c)/(b^4*d*x^4 + 4*a*b^3*d*x^3 + 6*a^2*b^2*d*x^2 + 4*a^3*b*d*x + a^4*d), x) - b*integrat
e(x/(b^4*d*x^4*e^(d*x + c) + 4*a*b^3*d*x^3*e^(d*x + c) + 6*a^2*b^2*d*x^2*e^(d*x + c) + 4*a^3*b*d*x*e^(d*x + c)
+ a^4*d*e^(d*x + c)), x) + 1/2*(x*e^(d*x + 2*c) - x*e^(-d*x))/(b^3*d*x^3*e^c + 3*a*b^2*d*x^2*e^c + 3*a^2*b*d*
x*e^c + a^3*d*e^c) - 1/2*a*e^(-c + a*d/b)*exp_integral_e(4, (b*x + a)*d/b)/((b*x + a)^3*b*d) + 1/2*a*e^(c - a*
d/b)*exp_integral_e(4, -(b*x + a)*d/b)/((b*x + a)^3*b*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\mathrm {cosh}\left (c+d\,x\right )}{{\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*cosh(c + d*x))/(a + b*x)^3,x)

[Out]

int((x*cosh(c + d*x))/(a + b*x)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(d*x+c)/(b*x+a)**3,x)

[Out]

Timed out

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